\(\int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [1062]

Optimal result

Integrand size = 50, antiderivative size = 124 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 C \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \] Output:

2*C*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^ 
(1/2))/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2*(B*b-2*C*a)*((a+b*cos(d*x+c))/(a 
+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*c 
os(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) C E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+(b B-2 a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{d \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + 
 b*Cos[c + d*x])^(3/2),x]
 

Output:

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((a + b)*C*EllipticE[(c + d*x)/2, (2 
*b)/(a + b)] + (b*B - 2*a*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(d*Sq 
rt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2014, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 2014

\(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{\sqrt {a+b \cos (c+d x)}}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+b^2 C \int \sqrt {a+b \cos (c+d x)}dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b^2 C \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 C \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 C \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^2 C \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {b^2 (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 C \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b^2 (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 C \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {2 b^2 (b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 C \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{b^2}\)

Input:

Int[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos 
[c + d*x])^(3/2),x]
 

Output:

((2*b^2*C*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/ 
(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*b^2*(b*B - 2*a*C)*Sqrt[(a + b* 
Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + 
b*Cos[c + d*x]]))/b^2
 

Defintions of rubi rules used

rule 2014

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S 
ymbol] :> Simp[1/b^2   Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], 
x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ 
[m, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.98

method	result	size
default	\(-\frac {2 \sqrt {\left (2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \left (B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b -2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a +C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b \right )}{\sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\)	\(246\)
parts	\(\text {Expression too large to display}\)	\(1125\)
risch	\(\text {Expression too large to display}\)	\(1383\)

Input:

int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/ 
2),x,method=_RETURNVERBOSE)
 

Output:

-2*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*(B*Elliptic 
F(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b-2*C*EllipticF(cos(1/2*d*x+1/2*c 
),(-2*b/(a-b))^(1/2))*a+C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) 
*a-C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)/(-2*b*sin(1/2*d*x 
+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2 
*d*x+1/2*c)^2*b+a+b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.99 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-3 i \, \sqrt {\frac {1}{2}} C b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 3 i \, \sqrt {\frac {1}{2}} C b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + \sqrt {\frac {1}{2}} {\left (-5 i \, C a + 3 i \, B b\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {\frac {1}{2}} {\left (5 i \, C a - 3 i \, B b\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )}}{3 \, b d} \] Input:

integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c 
))^(3/2),x, algorithm="fricas")
 

Output:

-2/3*(-3*I*sqrt(1/2)*C*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8 
/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8 
/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2* 
a)/b)) + 3*I*sqrt(1/2)*C*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, 
-8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, 
-8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 
2*a)/b)) + sqrt(1/2)*(-5*I*C*a + 3*I*B*b)*sqrt(b)*weierstrassPInverse(4/3* 
(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 
3*I*b*sin(d*x + c) + 2*a)/b) + sqrt(1/2)*(5*I*C*a - 3*I*B*b)*sqrt(b)*weier 
strassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*( 
3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b))/(b*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((B*a*b-a**2*C+b**2*B*cos(d*x+c)+b**2*C*cos(d*x+c)**2)/(a+b*cos(d 
*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c 
))^(3/2),x, algorithm="maxima")
 

Output:

integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/(b*c 
os(d*x + c) + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c 
))^(3/2),x, algorithm="giac")
 

Output:

integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/(b*c 
os(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {-C\,a^2+B\,a\,b+C\,b^2\,{\cos \left (c+d\,x\right )}^2+B\,b^2\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos(c + d*x))/(a + b*cos 
(c + d*x))^(3/2),x)
 

Output:

int((C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos(c + d*x))/(a + b*cos 
(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right ) b +a}d x \right ) a c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right ) b +a}d x \right ) b^{2}+\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \right ) b c \] Input:

int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/ 
2),x)
 

Output:

 - int(sqrt(cos(c + d*x)*b + a)/(cos(c + d*x)*b + a),x)*a*c + int(sqrt(cos 
(c + d*x)*b + a)/(cos(c + d*x)*b + a),x)*b**2 + int((sqrt(cos(c + d*x)*b + 
 a)*cos(c + d*x))/(cos(c + d*x)*b + a),x)*b*c